There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
- Each child must have at least one candy.
- Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
Example 1:
Input: [1,0,2] Output: 5 Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:
Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.
Approach #1: C++. [Using two array]
class Solution {
public:
int candy(vector<int>& ratings) {
int size = ratings.size();
int ans = 0;
vector<int> ltor(size, 1);
vector<int> rtol(size, 1);
for (int i = 1; i < size; ++i) {
int j = size - i - 1;
if (ratings[i] > ratings[i-1]) ltor[i] = ltor[i-1] + 1;
if (ratings[j] > ratings[j+1]) rtol[j] = rtol[j+1] + 1;
}
for (int i = 0; i < size; ++i)
ans += max(ltor[i], rtol[i]);
return ans;
}
};
Approach #2: Java. [Using one array]
public class Solution {
public int candy(int[] ratings) {
int[] candies = new int[ratings.length];
Arrays.fill(candies, 1);
for (int i = 1; i < ratings.length; i++) {
if (ratings[i] > ratings[i - 1]) {
candies[i] = candies[i - 1] + 1;
}
}
int sum = candies[ratings.length - 1];
for (int i = ratings.length - 2; i >= 0; i--) {
if (ratings[i] > ratings[i + 1]) {
candies[i] = Math.max(candies[i], candies[i + 1] + 1);
}
sum += candies[i];
}
return sum;
}
}
Approach #3: Java. [Single Pass Approach with Constant Space]
class Solution {
public int count(int n) {
return (n * (n + 1)) / 2;
}
public int candy(int[] ratings) {
if (ratings.length <= 1) return ratings.length;
int candies = 0;
int up = 0;
int down = 0;
int old_slope = 0;
for (int i = 1; i < ratings.length; ++i) {
int new_slope = ratings[i] > ratings[i-1] ? 1 : ratings[i] < ratings[i-1] ? -1 : 0;
if (old_slope > 0 && new_slope == 0 || old_slope < 0 && new_slope >= 0) {
candies += count(up) + count(down) + Math.max(up, down);
up = 0;
down = 0;
}
if (new_slope > 0) up++;
if (new_slope < 0) down++;
if (new_slope == 0) candies++;
old_slope = new_slope;
}
candies += count(up) + count(down) + Math.max(up, down) + 1;
return candies;
}
}
reference:
https://leetcode.com/problems/candy/solution/